原题
Assume that f(0) = 1 and 0^0=1. f(n) = (n%10)^f(n/10) for all n bigger than zero. Please calculate f(n)%m. (2 ≤ n , m ≤ 10^9, x^y means the y th power of x).
Input
The first line contains a single positive integer T. which is the number of test cases. T lines follows.Each case consists of one line containing two positive integers n and m.
Output
One integer indicating the value of f(n)%m.
Sample Input
2
24 20
25 20Sample Output
16
5解析
a^b mod c
可以转化为
a^(b mod phi(c)+phi(c)) mod c（b>phi(c)时成立）
所以要注意取模的时候 b<mod则不管，b>=mod时 b=b%mod+mod
AC代码
#include<bits/stdc++.h>
#include<cstring>
using namespace std;
typedef long long ll;
ll Oula(ll n)
{
ll ans=n;
for(int i=2;i<=(double)sqrt(n*1.0);i++)
{
if(n%i==0)
ans=ans*(1.0-1.0/i);
while(n%i==0)
n=n/i;
}
if(n>1)
ans=ans*(1.0-1.0/n);
return ans;
}
ll qpow(ll a, ll b, ll mod)
{
ll ans=1;
while(b)
{
if(b&1)
{
ans=ans*a;
if(ans>=mod)
ans=ans%mod+mod;
}
a*=a;
if(a>=mod)
a=a%mod+mod;
b>>=1;
}
return ans;
}
ll f(int n,int m){
if(n==0)
return 1;
if(n<10)
return n;
ll phi=Oula(m);
return qpow(n%10,f(n/10,phi),m);
}
int main(){
ll mod;
ll T,n;
cin>>T;
while(T--){
cin>>n>>mod;
cout<<f(n,mod)%mod<<endl;
}
}